Question : If $x=p+\frac{1}{p}$ and $y=p-\frac{1}{p}$, then the value of $x^{4}-2x^{2}y^{2}+y^{4}$ is:
Option 1: 24
Option 2: 4
Option 3: 16
Option 4: 8
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Correct Answer: 16
Solution :
$x = p+\frac{1}{p}$
$y = p-\frac{1}{p}$
Now, $x^{4}-2x^{2}y^{2}+y^{4}$
= $(x^{2}-y^{2})^{2}$
= $((x+y)(x-y))^{2}$
= $[(p+\frac{1}{p}+p-\frac{1}{p})(p+\frac{1}{p}-(p-\frac{1}{p})]^{2}$
= $[2×p×2×\frac{1}{p}]^{2}$
= 16
Hence, the correct answer is 16.
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